What is the integral of (9x^2-16x+4)/(x^3-3x^2+2x) ?

The integral of (9x^2-16x+4)/(x^3-3x^2+2x) is 3ln|x-1|+4ln|x-2|+2ln|x|+C

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integral of (9x^2-16x+4)/(x^3-3x^2+2x)

\int \frac{9 x ^{2} - 16 x + 4}{x ^{3} - 3 x ^{2} + 2 x} d x

3 ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣ + 2 ln ∣ x ∣ + C

Solution steps

\int \frac{9 x ^{2} - 16 x + 4}{x ^{3} - 3 x ^{2} + 2 x} d x

Expand

\frac{9 x ^{2} - 16 x + 4}{x ^{3} - 3 x ^{2} + 2 x} : \frac{9 x ^{2}}{x ^{3} - 3 x ^{2} + 2 x} - \frac{16 x}{x ^{3} - 3 x ^{2} + 2 x} + \frac{4}{x ^{3} - 3 x ^{2} + 2 x}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{9 x ^{2}}{x ^{3} - 3 x ^{2} + 2 x} d x - \int \frac{16 x}{x ^{3} - 3 x ^{2} + 2 x} d x + \int \frac{4}{x ^{3} - 3 x ^{2} + 2 x} d x

\int \frac{9 x ^{2}}{x ^{3} - 3 x ^{2} + 2 x} d x = 9 (- ln ∣ x - 1 ∣ + 2 ln ∣ x - 2 ∣)

\int \frac{16 x}{x ^{3} - 3 x ^{2} + 2 x} d x = 16 (- ln ∣ x - 1 ∣ + ln ∣ x - 2 ∣)

\int \frac{4}{x ^{3} - 3 x ^{2} + 2 x} d x = 4 (\frac{1}{2} ln ∣ x ∣ - ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 2 ∣)

= 9 (- ln ∣ x - 1 ∣ + 2 ln ∣ x - 2 ∣) - 16 (- ln ∣ x - 1 ∣ + ln ∣ x - 2 ∣) + 4 (\frac{1}{2} ln ∣ x ∣ - ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 2 ∣)

Simplify

9 (- ln ∣ x - 1 ∣ + 2 ln ∣ x - 2 ∣) - 16 (- ln ∣ x - 1 ∣ + ln ∣ x - 2 ∣) + 4 (\frac{1}{2} ln ∣ x ∣ - ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 2 ∣) : 3 ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣ + 2 ln ∣ x ∣

= 3 ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣ + 2 ln ∣ x ∣

Add a constant to the solution

= 3 ln ∣ x - 1 ∣ + 4 ln ∣ x - 2 ∣ + 2 ln ∣ x ∣ + C

\int sec (a θ) tan (a θ) d θ

y = 2, y = x^{2} - 2

\frac{\partial}{\partial y} (x^{\frac{1}{2}} y^{\frac{1}{2}})

12 x - x^{3}

x \to - 2 lim (3^{x + 1})

What is the integral of (9x^2-16x+4)/(x^3-3x^2+2x) ?
The integral of (9x^2-16x+4)/(x^3-3x^2+2x) is 3ln|x-1|+4ln|x-2|+2ln|x|+C